Let $g(x)=\dfrac{\sin(x)}{\sqrt{x}}$. $g'(x)=$
Solution: $g(x)$ is the quotient of two, more basic, expressions: $\sin(x)$ and $\sqrt{x}$. Therefore, the derivative of $g$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{\sin(x)}{\sqrt{x}}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sin(x))\sqrt{x}-\sin(x)\dfrac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\cos(x)\cdot \sqrt{x}-\sin(x)\cdot \dfrac1{2\sqrt{x}}}{(\sqrt{x})^2}&&\gray{\text{Differentiate }\sin(x)\text{ and }\sqrt{x}} \\\\ &=\dfrac{\sqrt{x}\cos(x)-\dfrac{\sin(x)}{2\sqrt{x}}}{x}&&\gray{\text{Simplify}} \\\\ &=\dfrac{2x\cos(x)-\sin(x)}{2x\sqrt{x}}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $g'(x)=\dfrac{2x\cos(x)-\sin(x)}{2x\sqrt{x}}$ or any other equivalent form.